Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $q = \dfrac{3y^3 + 18y^2 + 24y}{3y^3 + 3y^2 - 6y} \times \dfrac{y - 1}{-y + 5} $
First factor out any common factors. $q = \dfrac{3y(y^2 + 6y + 8)}{3y(y^2 + y - 2)} \times \dfrac{y - 1}{-(y - 5)} $ Then factor the quadratic expressions. $q = \dfrac {3y(y + 2)(y + 4)} {3y(y + 2)(y - 1)} \times \dfrac {y - 1} {-(y - 5)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { 3y(y + 2)(y + 4) \times (y - 1)} { 3y(y + 2)(y - 1) \times -(y - 5)} $ $q = \dfrac {3y(y + 2)(y + 4)(y - 1)} {-3y(y + 2)(y - 1)(y - 5)} $ Notice that $(y + 2)$ and $(y - 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {3y\cancel{(y + 2)}(y + 4)(y - 1)} {-3y\cancel{(y + 2)}(y - 1)(y - 5)} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $q = \dfrac {3y\cancel{(y + 2)}(y + 4)\cancel{(y - 1)}} {-3y\cancel{(y + 2)}\cancel{(y - 1)}(y - 5)} $ We are dividing by $y - 1$ , so $y - 1 \neq 0$ Therefore, $y \neq 1$ $q = \dfrac {3y(y + 4)} {-3y(y - 5)} $ $ q = \dfrac{-(y + 4)}{y - 5}; y \neq -2; y \neq 1 $